Having problems at calculating Eigenvectors, given complex Eigenvalues

Question

I want to find the Eigenvalues and Eigenvectors of the following matrix $$A=\begin{pmatrix}0&1\\-3&0\end{pmatrix}.$$

I'm able to calculate the Eigenvalues, which are simply the zero points of A's characteristic polynom $\lambda^2 + 3$:
$$\lambda_{1,2} = \pm i\sqrt{3}\\$$

But how do I calculate the Eigenvectors? I know that in theory the process is the same as in $\Re$ (clearly I lack some understanding even in $\Re$), but no matter what I do I cannot reach the same solution as is suggested by Wolframlpha: $$v_1=\begin{pmatrix}-i\sqrt{3}\\1\end{pmatrix}$$ $$v_2=\begin{pmatrix}+i\sqrt{3}\\1\end{pmatrix}$$

For example, when I try to solve the two equations \begin{equation} (i\sqrt{3})c_1+c_2 = 0 \\ -3c_1 + (i\sqrt{3})c_2 = 0 \end{equation}

I first multiply the upper row by $(-i\sqrt{3})$ which gives me $3c_1 - (-i\sqrt{3})c_2$. Adding this to the lower equation gives me $0=0$.

I've tried applying row operations as well (like here), but ultimately I end up at

$$\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

I would appreciate it if you could point my mistake out, or give me some advice on how to solve this problem.

Answer 1

Note that :

$$i\sqrt{3}c_1 + c_2 = 0 \Leftrightarrow c_2 = - i\sqrt{3}c_1$$

This means that you've already calculated that simply the coordinate $c_2$ of your eigenvector. This is an equation with infinite solutions, so just pick a simple value for $c_1$. Simply let $c_1 = 1$ and you will derive :

$$v_1=\begin{pmatrix}1\\-i\sqrt{3}\end{pmatrix}$$

Same goes exactly for the other eigenvector, $v_2$.

This is exactly what Wolfram Alpha gets (only it has divided by $\sqrt{3}$ so it just goes the either way around, both solutions of eigenvectors are correct since they span your desired eigenspace).

Your calculation and solution is correct. The $0=0$ happens because the two equations are actually one (the same, infinite solutions).

Answer 2

Note that $$\begin{pmatrix}0&1\\-3&0\end{pmatrix}\begin{pmatrix}1\\y\end{pmatrix}=\begin{pmatrix}y\\-3\end{pmatrix} $$ so you want $y=\lambda$ (and you also want $-3=\lambda y$).

Answer 3

I suggest checking your input to Wolfram|Alpha since neither vector that you’ve listed in your question is an eigenvector of $A$. For instance, $A[i\sqrt3,1]^T=[1,-3i\sqrt3]^T$, which is not a scalar multiple of $[i\sqrt3,1]^T$.

That aside, you should expect to end up with $0=0$ when you combine the two equations the way you did. Since $i\sqrt3$ is an eigenvalue of $A$, $A-i\sqrt3 I$ must be singular: its rows are linearly dependent, and so are the two equations that they represent—one is a multiple of the other. The system has an infinite number of solutions, and any non-zero solution will be an eigenvector.

The first equation says that $c_2=-i\sqrt3 c_1$. So, if you choose $c_1$ as the free variable, then any nonzero vector of the form $(c_1,-i\sqrt3 c_1)^T$ will be a corresponding eigenvector.